我使用Spring的WebServiceTemplate来使用Soap服务。有一段时间,这个Soap服务会响应一个无效的XML。我想拦截它的解析器并在解析之前修复无效的XML。我怎么能这样做?现在我打电话:
wsTemplate.sendSourceAndReceiveToResult(new StreamSource(new StringInputStream(msg)),new StreamResult(stringWriter))
我想我必须打电话给sendSourceAndReceive
并定义我自己的SourceExtractor
,但是当我真的想要做一些简单的事情时,我似乎过分干涉这个过程。
这是我想解决的问题:
An invalid XML character (Unicode: 0x1f) was found in the element content of the document.
SystemErr R at org.springframework.ws.soap.saaj.SaajSoapMessageFactory.createWebServiceMessage(SaajSoapMessageFactory.java :210)
该字符在XML 1.1中被接受,但该文档被描述为XML 1.0:
<?xml version="1.0" encoding="utf-8"?>
所以我想要的是为标签或空格替换该字符。
答案 0 :(得分:1)
我可以通过修改webServiceTemplate的WebServiceMessageFactory来解决之前修复过的消息,如下所示:
final WebServiceMessageFactory mfOriginal = myWebServiceTemplate.getMessageFactory();
WebServiceMessageFactory mfDecorator = new WebServiceMessageFactory() {
@Override
public WebServiceMessage createWebServiceMessage(final InputStream inputStream) throws InvalidXmlException, IOException {
InputStream decoratedIs = new InputStream() {
@Override
public int read() throws IOException {
int nextByte = inputStream.read();
// Replacing the invalid character with a tab. Touching nothing else.
if (nextByte == 0x1f) {
nextByte = 0x09;
}
return nextByte;
}
};
return mfOriginal.createWebServiceMessage(decoratedIs);
}
@Override
public WebServiceMessage createWebServiceMessage() {
return mfOriginal.createWebServiceMessage();
}
};
myServiceTemplate.setMessageFactory(mfDecorator);